Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gradient help

  1. Nov 3, 2003 #1
    Consider the function z=f(x,y). If you start at the point (4,5) and move toward the point (5,6), the direction derivative is sqrt(2). Starting at (4,5) and moving toward (6,6), the directional derivative is sqrt(5). Find gradient f at (4,5).

    Okay, this is probably a simple problem, but I don't know how to start it. Help appreciated.
  2. jcsd
  3. Nov 3, 2003 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Do you know a formula that relates gradients and directional derivatives?

    What do you get when you plug what you know into that formula? (represent unknowns with variables)
  4. Nov 3, 2003 #3
    Forumula...*scratches head*

    Um, I think this is the one, hopefully:
    The derivative of f at Po in the direction of u= gradient f dot with u = magnitude grad f * magnitude u * cos theta = magnitude grad f * cos theta.

    Or like D_u f= grad f (dot) u = |grad f| |u| cos theta = |grad f| cos theta

    Maybe that'll help me out?
  5. Nov 4, 2003 #4


    User Avatar
    Science Advisor

    Yes, that's true as long as u is a unit vector (has length 1).

    Another way of saying this is that the directional derivative is
    f_x u_x+ f_y u_y where f_x is the partial derivative and u_x is the x component of unit vector u.

    Find unit vectors in both the given directions and write out
    f_x u_x+ f_y u_y= √(2) and
    f_x v_x+ f_y v_y= √(5).

    That gives you two equations for f_x and f_y.
  6. Nov 4, 2003 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That's the one of which I was thinking.

    You know two instances of D_u f and u. (don't forget that u is a unit vector!), so grad f is the unknown. You have two equations, so you should be able to solve for the two components of grad f.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook